Question

Find a polynomial whose zeros are square of the zeros of polynomial 3 x square + 6 x minus 9

Answer 1

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Answer 2

Answer:

$x^2 - (\frac{8}{3} )x + (\frac{4}{3})= 0$

Step-by-step explanation:

Given : Polynomial $3x^2+6x-9$

To find : The polynomial whose zeros are reciprocal of the zeros of the given polynomial.

Solution : First we find the roots of the given polynomial.

Equate the given polynomial to zero to find the roots

$3x^2+6x-9=0$

Apply Discriminant and the solution is in the form

$x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

where a= 3 , b=6 , c=-9

$x= \frac{-6\pm\sqrt{6^2-4(3)(-9)}}{2(3)}$

$x= \frac{-6\pm\sqrt{36+108}}{6}$

$x= \frac{-6\pm\sqrt{144}}{6}$

$x= -1\pm2$

Therefore, $x=1$ , $x=-3$

Reciprocal of Roots are  $x=1$ , $x= \frac{-1}{3}$

To form a polynomial with roots 1 + α and 1 + β. This yields the equation,

$\alpha= 1$ , $\beta= \frac{-1}{3}$

New equation = $(x-(1+\alpha))(x-(1+\beta )$

$x^2 - (2 +\alpha + \beta )x + (1 + \alpha )(1 + \beta ) = 0$

Put value of α and β

$x^2 - (2 +(1) + ( \frac{-1}{3}) )x + (1 +(1) )(1 + ( \frac{-1}{3}))= 0$

$x^2 - (\frac{8}{3} )x + (\frac{4}{3})= 0$

This is the new equation.